PG 25 4 Apr 2019Easter challenges

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteaux word combining parallel and telegram. A message sent each week by the Parallel Project to bright young mathematicians.
  • Tackle each Parallelogram in one go. Don’t get distracted.
  • Finish before the end of Easter holidays.
  • Your score & answer sheet will appear immediately after you hit SUBMIT.
  • Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.

Junior Maths Challenge (UKMT) Mock

Some of you will be doing the UKMT’s Junior Maths Challenge on 30th April, so this Parallelogram consists of 25 Junior Maths Challenge questions – some easy, some tricky. This is an opportunity to get some practice before the real JMC. And, if you are not doing the JMC, then it is an excellent way to develop your maths skills.

The real JMC allows you one hour to attempt 25 questions, so in theory you have roughly sixty minutes to complete these 25 questions. However, don’t rush, don’t guess and don’t worry if you need longer than sixty minutes.

Here are a few things to bear in mind when you do the real JMC:

  • you may not use a calculator.
  • the first half of the questions are not too tricky (also true below).
  • the second half of the questions are really tricky (also true below).
  • you will lose marks for wrong answers in the second half of the questions (not true below).

So don’t take wild guesses for the second half of the questions in the real JMC, and don’t take wild guesses for trickier questions below. However, if you have a rough idea of the answer or have narrowed it down to a couple of possible answers, then it is worth picking an answer.

Make sure you have a paper and pencil to hand for sketches and calculations.

Because this Parallelogram is longer and tougher than a typical Parallelogram, we will be awarding TREBLE POINTS for this week's answers! So there's even more reason to give this one your best shot.

Finally, afterwards, make sure you go through all the answers and see where you went right and where you went wrong. This is 100% the best way to improve your maths skills.

The next Parallelogram will contain our usual mix of videos, humour, puzzles and problems.

1.

5 marks

1.1 What is half of 1.01?

  • 5.5
  • 0.55
  • 0.505
  • 0.5005
  • 0.055

We obtain half of 1.01 by dividing 1.01 by 2. We can do this as a long division sum, or alternatively, we can use fractions:

1.01=1+1100. So half of 1.01 is 12+1200=0.5+0.005=0.505

2.

5 marks

2.1 Beatrix looks at the word JUNIOR in a mirror. How many of the reflected letters never look the same as the original, no matter how Beatrix holds the mirror?

  • 1
  • 2
  • 3
  • 4
  • 5

The letters J, N and R do not have an axis of symmetry. So these letters cannot look the same when reflected in a mirror, however the mirror is held. The letters U, I and O all have at least one axis of symmetry. So each may look the same when reflected in a mirror.

3.

5 marks

3.1 The diagrams on the right show three different views of the same cube. Which letter is on the face opposite U?

  • I
  • P
  • K
  • M
  • O

There are two approaches.

(A) Draw a net of the cube, and straight away the middle diagram allows us to fill in the large letters. If U is in the middle, then the square at the bottom of the net is opposite U. Then, we can look at the first diagram, which gives us the possibilities in small letters, because I and M must be next to K. The third diagram demands that M is next to O, so M must be in the bottom square of the net, so M is opposite U.

(B) We see from the left hand diagram that K is not opposite either I or M, and from the middle diagram that it is not opposite either O or U. Therefore K must be opposite P. So neither K nor P is opposite U. From the middle diagram O is also not opposite U. So U is opposite either I or M. But if U is opposite I, then O must be opposite M, and this possibility is ruled out by the right hand diagram. So U must be opposite to M (and, also, I is opposite O).

4.

5 marks

4.1 Tommy Thomas's tankard holds 480 ml when it is one quarter empty. How much does it hold when it is one quarter full?

  • 120 ml
  • 160 ml
  • 240 ml
  • 960 ml
  • 1440 ml

When Tommy's tankard is one quarter empty it is three quarters full. So 480 ml is three quarters of the capacity of the tankard. So when it is one quarter full it holds 13480=160 ml.

5.

5 marks

5.1 The diagram shows two arrows drawn on separate 4 cm × 4 cm grids. One arrow points North and the other points West.

When the two arrows are drawn on the same 4 cm × 4 cm grid (still pointing North and West) they overlap. What is the area of overlap?

  • 4 cm2
  • 4 12 cm2
  • 5 cm2
  • 5 12 cm2
  • 6 cm2

By drawing one arrow on top of the other, as shown, we see that the region of overlap covers the whole of 4 of the 1 cm × 1 cm squares into which the grid is divided, and 4 halves of these squares. So the area of the overlapping region is 4 + 4 (12) = 6 cm2.

6.

5 marks

6.1 Laura wishes to cut this shape, which is made up of nine small squares, into pieces that she can then rearrange to make a 3×3 square.

What is the smallest number of pieces that she needs to cut the shape into so that she can do this?

  • 2
  • 3
  • 4
  • 5
  • 6

In a 3×3 square each row and column contains just 3 squares. So none of the pieces that Laura uses to make the square can be more than 3 squares long. It follows that the squares labelled a and b must be in different pieces, as must be the squares labelled b and c. So there must be at least three pieces. The diagrams below show how Laura can fulfill her task using 3 pieces.

7.

5 marks

7.1 A pattern that repeats every six symbols starts as shown below:

♥ ♣ ♥ ♦ ♥ ♠ ♥ ♣ ♥ ♦ ♥ ♠ ...

Which are the 100th and 101st symbols, in that order, in the pattern?

  • ♦ ♥
  • ♥ ♦
  • ♥ ♣
  • ♠ ♥
  • ♣ ♥

The pattern repeats every 6 symbols. Therefore, as 96 is a multiple of 6, the symbols in positions 97, 98, 99, 100, 101, 102, are the same as those in positions 1, 2, 3, 4, 5, 6, that is, they are ♥ ♣ ♥ ♦ ♥ ♠. We see from this that the symbols that are the 100th and 101st in list are ♦ ♥, in this order.

Can you see the link with modular arithmetic, something we’ve covered in previous Parallelograms?

8.

5 marks

8.1 Which of the following calculations gives the largest answer?

  • 2 − 1
  • 2 ÷ 1
  • 2 × 1
  • 1 × 2
  • 2 + 1

We have

2 − 1 = 1,
2 ÷ 1 = 2,
2 × 1 = 2,
1 × 2 = 2,

and

2 + 1 = 3.

We therefore see that 2 + 1 gives the largest value.

9.

5 marks

9.1 What is the value of x ?

  • 43
  • 47
  • 53
  • 57
  • 67

Let the angle shown be y°.

Because the angles at a point total 360°, we have y + 303 = 360.

Therefore y = 360 − 303 = 57.

Because the alternate angles formed by a line which cuts a pair of parallel lines are equal, x=y.

Hence x = 57.

10.

5 marks

10.1 What is the value of 201 × 7 − 7 × 102 ?

  • 142 800
  • 793
  • 693
  • 607
  • 0

We have 201 × 7 − 7 × 102 = 1407 − 714 = 693.

This straightforward method does not involve too much hard work. However, as both the products 201 × 7 and 7 × 102 involve the factor 7, there is an alternative method in which we first take out this common factor:

201 × 7 − 7 × 102
= 7 × (201 − 102)
= 7 × 99
= 7 × (100 − 1)
= 7 × 100 − 7 × 1
= 700 − 7
= 693.

11.

5 marks

11.1 If you work out the values of the following expressions and then place them in increasing numerical order, which comes in the middle?

  • 23+45
  • 23×45
  • 32+54
  • 23÷45
  • 32×54

We evaluate each expression in turn, and we write the answers as fractions with the same denominator so as to make it easy to arrange them in increasing order.

We have:

23+45=10+1215=2215=176120

23×45=2×43×5=815=64120

32+54=12+108=228=330120

23÷45=23×54=1012=100120

32×54=3×52×4=158=225120

We now see that when the fractions are arranged in increasing numerical order, the order is:

23×45, 23÷45, 23+45, 32×54, 32+54

Therefore it is the expression

23+45

whose value is in the middle.

12.

5 marks

12.1 In William Shakespeare’s play As You Like It, Rosalind speaks to Orlando about, "He that will divide a minute into a thousand parts".

Which of the following is equal to the number of seconds in one thousandth of one minute?

  • 0.24
  • 0.6
  • 0.024
  • 0.06
  • 0.006

There are 60 seconds in one minute. Therefore the number of seconds in one thousandth of a minute is:

601000=6100=0.06.

13.

5 marks

13.1 Seven squares are drawn on the sides of a heptagon so that they are outside the heptagon, as shown in the diagram.

What is the sum of the seven marked angles?

  • 315°
  • 360°
  • 420°
  • 450°
  • 630°

Method 1: In the figure we have labelled some of the vertices so that we may refer to them.

Suppose that there is a flag whose pole is in the direction of GP and pointing as shown. Consider the effect of carrying out the following operations. First rotate the flag anti-clockwise about G through PGQ so that now its pole lies along GQ. Next slide the flag without rotation so that its pole lies along HR. Next rotate the pole about H through RHS so that it lies along HS, and so on, until the flag returns to its original position.

The total angle that the flag has turned through is the sum of the seven marked angles. But in returning to its original position the flag has completed a full rotation of 360°.

Therefore the sum of the seven marked angles is 360°.

Method 2: We use the same labelling of the vertices that we used in Method 1. We let the sum of the marked angles be X°.

The sum of the angles at the vertex G is 360°. Now FGPand∠QGH are angles of a square, and so are each 90°. It follows that PGQ+HGF=360°90°90°=180°. Therefore

PGQ=180°HGF.

A similar equation holds for each of the seven marked angles. It follows, by adding the seven equations obtained in this way, that

X°=7×180° − the sum of the interior angles of the heptagon.

The sum of the interior angles of a polygon with n vertices is n2×180°. Therefore the sum of the interior angles of a heptagon is 5×180°. Therefore

X°=7×180°5×180°=2×180°=360°.

14.

5 marks

14.1 Consider the following three statements.

  • (i) Doubling a positive number always makes it larger.
  • (ii) Squaring a positive number always makes it larger.
  • (iii) Taking the positive square root of a number always makes it smaller.

Which statements are true?

  • All three
  • None
  • Only (i)
  • (i) and (ii)
  • (ii) and (iii)

Statement (i) is true, because, if x is a positive number, then

2x=x+x>x.

Statement (ii) is false. If a positive number which is less than 1 is squared, then the answer is smaller than the original number. For example, with x=12, we have:

x2=14<12.

Similarly, if we take the positive square root of a number that is less than 1, then the answer is larger than the original number. For example

14=12>14.

Hence statement (iii) is false.

Therefore statement (i) is the only one that is true.

15.

5 marks

15.1 What is the remainder when the square of 49 is divided by the square root of 49?

  • 0
  • 2
  • 3
  • 4
  • 7

Because 49=7×7, it follows both that 49=7 and that 7 is a factor of 49.

Therefore 7 is also a factor of 492.

Hence the remainder when the square of 49 is divided by the square root of 49 is 0.

Next up, we have the trickier questions.

16.

6 marks

16.1 Only one of the following statements is true. Which one?

  • A ‘B is true’
  • B ‘E is false’
  • C ‘Statements A to E are true’
  • D ‘Statements A to E are false’
  • E ‘A is false’

For statement A to be true, B would also have to be true. But we are told that only one statement is true, so this is not possible. We deduce that statement A is false. Therefore statement E is true.

In the context of the JMC you could stop here, because we have found one statement that must be true. However, we really ought to satisfy ourselves that statement E is the only one that is true.

As statement E is true, statement B is false. Statement D can never be true, because if statement D were true, all the statements, including D, would be false, and we would have a contradiction. Because statement D must be false, statement C cannot be true.

So statement E is true and all the other statements are false.

17.

6 marks

17.1 Beth, Carolyn and George love reading their favourite bedtime stories together. They take it in turns to read a page, always in the order Beth, then Carolyn, then George. All twenty pages of the story are read on each occasion. One evening, Beth is staying at Grandma's house but Carolyn and George still read the same bedtime story and take it in turns to read a page with Carolyn reading the first page.

In total, how many pages that Carolyn and George read are the same as the pages that they would read if Beth was there?

  • 1
  • 2
  • 4
  • 6
  • 7

One way to do this is to write down a list of 20 pages and then label how they would be read usually and then on the night that Beth is away.

Usual B C G B C G B C G B
Page 1 2 3 4 5 6 7 8 9 10
Beth C G C G C G C G C G
Usual C G B C G B C G B C
Page 11 12 13 14 15 16 17 18 19 20
Beth C G C G C G C G C G

By comparison, you can see that pages 5, 6, 11, 12, 17 & 18 are read by the same person, regardless of Beth’s absence. Next is a more mathematical approach to the problem.

When all three children are present, Carolyn reads pages 2, 5, 8, ... , that is those pages whose number leaves remainder 2 when divided by 3. George reads pages 3, 6, 9, ... that is, those whose number is a multiple of 3. When Beth stays at Grandma's, Carolyn reads pages 1, 3, 5, ... that is the odd numbered pages, and George reads the even numbered pages. So the pages that Carolyn reads both normally and also when Beth is away are those with numbers which have remainder 2 when divided by 3, and are odd. That is, the three pages 5, 11 and 17.

The pages that George reads both normally and also when Beth is away are those with numbers that are multiples of 3 and are even, that is, the three pages 6, 12 and 18. So there are altogether 6 pages which are read by the same person normally and when Beth is away, namely pages 5, 6, 11, 12, 17 and 18.

18.

6 marks

18.1 Which of these statements is true?

  • 15614=1+561×4
  • 15615=1+561×5
  • 15616=1+561×6
  • 15617=1+561×7
  • 15618=1+561×8

In the context of the JMC we can assume that just one of the given options is correct, so we can find which it is by eliminating the ones that are wrong. We can do this by just considering the last digit (the units digit) of the given options.

The last digit of 56 is 5. Since 1+561×4=1+564 its last digit is the same as that of 1 + 5 − 4, that is, 2. So the first option is not the correct answer.

In a similar way, it follows that the last digit of 1+561×5 is 1, the last digit of 1+561×6 is 0, and the last digit of 1+561×7 is 9.

So this leaves 15618=1+561×8 as the only possible correct option.

In this way there is no need to evaluate 56. However, to give a complete answer we would need to check our answer is correct. This is straightforward:

1+561×8=1+156258=156268=15618.

19.

6 marks

19.1 The numbers 2, 3, 4, 5, 6, 7, 8 are to be placed, one per square, in the diagram shown such that the four numbers in the horizontal row add up to 21 and the four numbers in the vertical column add up to 21.

Which number should replace x?

  • 2
  • 3
  • 5
  • 7
  • 8

If we add all the numbers in the horizontal column and all the numbers in the vertical row, we get a total of 21 + 21 = 42 . In doing this sum we add in all the numbers 2, 3, 4, 5, 6, 7, 8 once except for x which is added in twice. So the total we get is 2 + 3 + 4 + 5 + 6 + 7 + 8 + x = 35 + x. Since this equals 42, we must have x = 7.

To complete the solution we should check that with x = 7, it is possible to place the remaining numbers in the other squares so that the four numbers in the horizontal row add up to 21, and so also do the four numbers in the vertical column.

20.

6 marks

20.1 Box P has p chocolates and box Q has q chocolates, where p and q are both odd and p>q.

What is the smallest number of chocolates which would have to be moved from box P to box Q so that box Q has more chocolates than box P?

  • qp+22
  • pq+22
  • q+p22
  • pq22
  • q+p+22

Because p and q are both odd, p+q is even. If the chocolates were shared equally between the boxes there would be 12p+q chocolates in each box.

So the least number of chocolates there must be in box Q if it is to have more chocolates than box P is 12p+q+1.

Because box Q starts with q chocolates in it, to end up with 12p+q+1 in box Q the number of chocolates we need to transfer from box P to box Q is

p+q2+1q=pq2+1=pq+22

21.

6 marks

21.1 Aaron says his age is 50 years, 50 months, 50 weeks and 50 days old. What age will he be on his next birthday?

  • 56
  • 55
  • 54
  • 53
  • 51

50 months is 4 years and 2 months. 50 weeks is around 11 12 months and 50 days is about 1 23 months.

So Aaron's age is approximately (50 + 4) years + (2 + 11 12 + 1 23) months = 54 years + 15 16 months = 55 years and 3 16 months. So he will be 56 on his next birthday.

22.

6 marks

22.1 In the division calculation 952,473 ÷ 18, which two adjacent digits should be swapped in order to increase the result by 100?

  • 9 and 5
  • 5 and 2
  • 2 and 4
  • 4 and 7
  • 7 and 3

If after division by 18 the number has to be increased by 100, then before the division it needs to be increased by 18 × 100 = 1800. So we need to find two adjacent digits in the number 952,473 which when swapped increase it by 1800.

If 952,473 is increased by 1800 it becomes 952,473 + 1 800 = 954,273.

We obtain 954,273 from 952,473 by swapping the adjacent digits 2 and 4.

23.

6 marks

23.1 The diagram shows a design formed by drawing six lines in a regular hexagon. The lines divide each edge of the hexagon into three equal parts.

What fraction of the hexagon is shaded?

  • 15
  • 29
  • 14
  • 310
  • 516

If we draw in the additional lines shown in the diagram on the right, the hexagon is divided into 54 small congruent equilateral triangles of which 12 are shaded. So the fraction that is shaded is 1254 = 29.

In fact, it is easier to note that the hexagon can be divided into 6 congruent equilateral triangles, like the one shown with the bold edges.

Each of these is made up of 9 of the small equilateral triangles of which 2 are shaded.

24.

6 marks

24.1 The diagram shows a regular octagon with sides of length 1.

The octagon is divided into regions by four diagonals.

What is the difference between the area of the hatched region and the area of the region shaded grey?

  • 0
  • 18
  • 14
  • 12
  • 1

The octagon is divided into the central square, which has sides of length 1, four congruent rectangles and four congruent right-angled triangles. The hatched region is made up of two of the rectangles, and the square. The grey area is made up of two of the rectangles and three of the triangles. Therefore the difference between their areas is the difference between the area of the square and the area of three of the triangles.

The square has area 1. Each of the triangles is a right-angled isosceles triangle with hypotenuse of length 1. Let x be the length of each of the other two sides. By Pythagoras’ Theorem, x2+x2=12, that is 2x2=1 and so x2=12. So the area of each triangle is 12x2=14.

Therefore the difference between the area of the square and the area of three of the triangles is 1314=14.

We can also see this geometrically. The diagram on the right shows that four of the triangles fit together to make a square whose side has length 1. So the difference between the area of the square and that of three of the triangles is the area of one-quarter of the square.

25.

6 marks

25.1 After playing 500 games, my success rate at Spider Solitaire is 49%. Assuming that I win every game from now on, how many extra games do I need to play in order that my success rate increases to 50%?

  • 1
  • 2
  • 5
  • 10
  • 50

Since I have won 49% of my first 500 games, so far I have won 49100×500=49×5=245 games. So I have lost 500245=255 games. I need now to win enough games so that I have won as many as I have lost.

So, assuming I win every game from now on, I need to win 255245=10 more games.

Another approach is to say that I am going play and win W more games. At that point, the number of games that I win will be:

49100×500+W500+W=50%or0.5

Therefore 245+W500+W=0.5

245+W=250+0.5W
0.5W=5
W=10

Before you hit the SUBMIT button, here are some quick reminders:

  • You will receive your score immediately, and collect your reward points.
  • You might earn a new badge... if not, then maybe next week.
  • Make sure you go through the solution sheet – it is massively important.
  • A score of less than 50% is ok – it means you can learn lots from your mistakes.
  • The next Parallelogram is after the Easter holidays.
  • Finally, if you missed any earlier Parallelograms, make sure you go back and complete them. You can still earn reward points and badges by completing missed Parallelograms.

Cheerio, Simon.